Question

The total cost of producing \['x'\]radio sets per day is Rs. \[\left( \dfrac{{{x}^{2}}}{4}+35x+25 \right)\]and price per set at which they may be sold is Rs. \[\left( 50-\dfrac{x}{2} \right)\]. Find the daily output to maximise the total profit.

Answer 1

Hint: We solve this problem first by calculating the profit. If C.P is the cost price and S.P is the selling price of \['x'\]radio sets per day, then the profit is calculated by using the formula
 \[P=S.P-C.P\]
Here, as P is the function of \['x'\], to find the output of maximum profit we solve the value of \['x'\]by making the differentiation of P to zero that is
 \[\Rightarrow \dfrac{d}{dx}\left( P \right)=0\]
The value of \['x'\]from the above equation will give the required answer.

Complete step-by-step answer:
We are given that the cost price of \['x'\]radio sets per day is
 \[\Rightarrow C.P=\left( \dfrac{{{x}^{2}}}{4}+35x+25 \right)\]
We are given that selling price of one radio set is \[\left( 50-\dfrac{x}{2} \right)\]
Now, let us calculate the total selling price of \['x'\]radio sets per day as follows
 \[\begin{align}
  & \Rightarrow S.P=x\left( 50-\dfrac{x}{2} \right) \\
 & \Rightarrow S.P=50x-\dfrac{{{x}^{2}}}{2} \\
\end{align}\]

We know that the profit is calculated by using the formula \[P=S.P-C.P\].
By using the above formula we get the profit as
 \[\Rightarrow P=\left( 50x-\dfrac{{{x}^{2}}}{2} \right)-\left( \dfrac{{{x}^{2}}}{4}+35x+25 \right)\]
By dividing the terms of same power we get
 \[\begin{align}
  & \Rightarrow P=\left( -\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{2}}}{4} \right)+\left( 50x-35x \right)-25 \\
 & \Rightarrow P=-\dfrac{3{{x}^{2}}}{4}+15x-25 \\
\end{align}\]
Now, let us find the value of \['x'\]for which the profit P is maximum.
We know that, to find the value of \['x'\]for which the function is the maximum is calculated by solving the equation
 \[\Rightarrow \dfrac{d}{dx}\left( P \right)=0\]
By substituting the function P in above equation we get
 \[\Rightarrow \dfrac{d}{dx}\left( -\dfrac{3{{x}^{2}}}{4}+15x-25 \right)=0\]
Now, by dividing the terms we get
 \[\begin{align}
  & \Rightarrow -\dfrac{3}{4}\left( \dfrac{d}{dx}\left( {{x}^{2}} \right) \right)+15\dfrac{d}{dx}\left( x \right)-25\dfrac{d}{dx}\left( 1 \right)=0 \\
 & \Rightarrow -\dfrac{3x}{2}+15=0 \\
 & \Rightarrow x=\dfrac{15\times 2}{3}=10 \\
\end{align}\]
Therefore, we can say that the profit is the maximum when there is a daily output of 10 radio sets.

Note: Students will make mistakes in selling the price of the radio sets. We are given the selling price of one radio set as \[\left( 50-\dfrac{x}{2} \right)\]. But students will consider this price as the selling price of \['x'\]radio sets which will be wrong. They consider \[S.P=\left( 50-\dfrac{x}{2} \right)\]which leads to wrong answers. So, reading the question and considering the values is important.